Answer
$\eta_{pump}=91.79%$
Work Step by Step
Here we have a change of pressure and velocity then:
$\Delta E_{fluid}=Q\Delta P+\frac{1}{2}m\Delta V^2=Q*(\Delta P+\frac{1}{2}\rho \Delta V^2)$
First we calculate the areas of the entrance (1) and the exit (2):
$A_{1}=\frac{\pi D_{1}^2}{4}=\frac{\pi*(0.08m)^2}{4}=0.005027m^2$
$A_{2}=\frac{\pi D_{2}^2}{4}=\frac{\pi*(0.12m)^2}{4}=0.011310m^2$
Now we calculate the velocities at the entrance (1) and the exit (2) of the pump:
$V_{1}=\frac{Q}{A_{1}}=\frac{0.1\frac{m^3}{s}}{0.005027m^2}=19.89\frac{m}{s}$
$V_{2}=\frac{Q}{A_{2}}=\frac{0.1\frac{m^3}{s}}{0.011310m^2}=8.84\frac{m}{s}$
Then:
$\Delta E_{fluid}=0.1\frac{m^3}{s}*(500kPa+\frac{1}{2}*860\frac{kg}{m^3}*((8.84\frac{m}{s})^2-(19.89\frac{m}{s})^2)$
$\Delta E_{fluid}=36.35kW$
The mechanical energy suministrated is:
$E_{m}=\eta E_{e}=0.90*44kW=39.6kW$
Then the efficiency of the pump is:
$\eta_{pump}=\frac{E_{fluid}}{E_{m}}=\frac{36.35kW}{39.6kW}=0.9179$
$\eta_{pump}=91.79%$
